Section C: Others

In section C is participated 9 entries from 9 authors from 6 countries.

List of entries:

Nr.

Author

Country

Problems

Count

1

Bandžuch Imrich

SVK

h#2, h#5,5

2

2

Bourd Evgeni

IZR

ser-x10

1

3

Coakley Jeff

CAN

*mathematic

0,5

4

Denkovski Ivan

MAC

PG 19,5

1

5

Frolkin Andrey

UKR

*mathematic

0,5

6

Lörinc Juraj

SVK

mathematic

1

7

Packa Ladislav

SVK

mathematic

1

8

Skoba Ivan

CZE

h#5

1

9

Storisteanu Adrian

CAN

mathematic

1

* co-autor

A rest nine problems I have assigned in this section. Because all problems are very specific I have decided to award each of them.

1st prize| section C

Juraj Lörinc

I. cena

Marián Križovenský 55 JT (C)
C 4.4.2016

No animated solution!

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White king visits in series of moves all chessboard corners and returns to h1 in the shortest possible time.

How many solutions?

[C6]

See text! → (3+3)

Detailed solution:
The distance between any two corners is the same, 7 moves. I.e. the theoretical minimal length of series is 28. Is such minimal roundtrip possible?

The join h1↔a8 is excluded as Pb7 and Ba6 guard each other. Also the route in direction a1→a8 is excluded as Sa8 and Pb7 guard squares a6, b6, c6, included in each shortest path a1→a8. But the opposite direction a8→a1 is allowed as Sa8 will be captured first, only then wK will pass through b6.

Thus a8 must be entered from h8 and bK must continue to a1. As a consequence we have formed the only possible sequence of corners h1→h8→a8→a1→h1, composed of 4 segments. Example of one specific journey:

Each of four segments of the king’s journey is independent of other, thus any specific journey in one segment is possible to combine with any journeys in other segments. As a consequence, it is enough to find number of ways from one corner to the following and then multiply these numbers.

How is it possible to find number of ways from corner to corner? For two neighbouring corners it is sufficient to realize the following principles, resulting in the recursive calculation. We will show the on the movement from corner (0,0) – e.g. a1 – to corner (7,0) – e.g. h1. We will denote number of ways to square (x,y) as P(x,y).

There is exactly 1 possibility how the king can move in 0 moves from corner (0,0) to the same square, i.e. P(0,0) = 1.

For squares that cannot be entered by king while moved from corner to corner and (for x < 0 or y < 0), i.e. for coordinates out of board, P(x,y) = 0.

For all other squares while moving in direction from (0.0) to (7,0), the recursive formula holds:
P(x,y) = P(x-1, y-1) + P(x-1, y) + P(x-1, y+1).

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a

2a+b

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a+b

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4a+4b+2c

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a+b+c

2a+3b+2c

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[Tabuľka 1]

(roh1 → roh2)

The following table demonstrates these rules with some examples (red cell denotes square that cannot be visited by the king).

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21

51

127

[Tabuľka 2]

(roh1 → roh2)

In the case of the empty chessboard, the number of the shortest ways from corner to corner is 127 for neighbouring corners (as shown in the following table, empty cells correspond to zeros, as there is no even theoretical shortest way via them) and 1 for opposite corners (but this was already excluded in the present problem).

Let’s have a look now on the individual segments of the proposed journey of wK in the form of tables.

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91

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56

35

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[Schéma 1]

(h1→h8)

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[Schéma 2]

(h8→a8)

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[Schéma 3]

(a8→a1)

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37

[Schéma 4]

(a1→h1)

As the last step of solution it is necessary to multiply results for individual segments, to get the total number of solutions:
91 x 11 x 15 x 37 = 555 555.

Comment:
Author sent me this problem first without solution to try solve it myself. After about two hours I have concluded I have found a solution algorithm and I estimated outcome as 5555. Wenn I then looked at authors solution, I find that my appraisal was exactly 100 times smaller.
Perfect symbiosis of author’s background as mathematician as well as chess composer. It is here more problems with similary stipulation of count of solution but outcome of this problem is amazing. Doubtless deserved first prize!

Remark:
In diagramm C6b I created animation of one solution. For all 555 555 solutions is also internet to small and human life to short…

2nd prize| section C

Evgeni Bourd

II. cena

Marián Križovenský 55 JT (C)
C 4.4.2016

No animated solution!

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White does all moves and needs to capture in 5+5 moves.No short solutions.

Comment:
Interesting condition for count of solutions but injury it needs auxiliary condition “No short solutions”. There is one single 9-move solution. But pieces arrangement presents symbol 55 what is thematic item too.
Original, thematic, economic!

3rd prize| section C

Adrian Storisteanu

III. cena

Marián Križovenský 55 JT (C)
C 4.4.2016

No animated solution!

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a)Add QQQQ for a symmetrical position of five guarded pieces with 55 moves

The wQ already provided in the diagram ensures a unique solution, out of the possible rotated and reflected settings. A better alternative to the term ‘guarded’ (or ‘defended’), commonly used in such construction tasks, might be ‘observed’ (especially given the presence of a K in twin b).

Solution

Adrian Storisteanu

III. cena

Marián Križovenský 55 JT (C)
C 4.4.2016

No animated solution!

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a)There are 18 base positions (not counting the usual rotations and reflections, that is) of 5 guarded Qs with 55 available moves on a 5x5 board. Only one setting is symmetrical. The given wQ in the diagram is not on the axis of symmetry of the final position, which may complicate a potential solver’s attempts!?...

[C8b]

Solution a) (5+0)

Adrian Storisteanu

III. cena

Marián Križovenský 55 JT (C)
C 4.4.2016

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b)There is one base position of unguarded K + 4Qs that have a total of 55 possible moves on a 5x5 board.

[C8c]

Solution b) (5+0)

Comment:
Five-level variations on famous problem of eight queens. A bit breakneck additional conditions but result is noteworthy. Only by solution must be man more mathematician as composer.

1st honourable mention| section C

Andrey FrolkinJeff Coakley

1. čestné uznanie

Marián Križovenský 55 JT (C)
C 4.4.2016

No animated solution!

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Each letter represents a different type of piece.Upper case is one color, lower case is the other.

Determine the position.

[C5a]

See the text → (5+5)

Solution

Andrey FrolkinJeff Coakley

1. čestné uznanie

Marián Križovenský 55 JT (C)
C 4.4.2016

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There are five pairs of letters on the board: F/f I/i V/v E/e S/s ('fives'). Of these, only I/i occupy non-adjacent squares. This means that they are kings. If v is a queen or a bishop, the king on e2 is in illegal check from vd1. Thus two options remain for V/v: knight or rook. If this letter stands for a knight, E/e can only represent a bishop; otherwise Ke2 is in illegal check from Ee1. The remaining vacancies are queen and rook; in both cases (F/f = queen, S/s = rook or vice versa) the kings are in check simultaneously, which is impossible. Therefore, V/v = rook. The king on c2 is in check; if e or s represents a queen, the other king is also in check; hence one of these letters is a knight and the other is a bishop. If Ee1 is a knight, then Kc2 is in illegal double check from Rc1 and Se1. So E/e = knight and S/s = bishop. The only possibility remaining for F/f is to represent the queens. Again the king on c2 is in double check, but this time it is legal because last move must have been b2xR/B/Sc1=R++!

[C5b]

Solution → (5+5)

Comment:
This is something for crosswords lovers.
Their favorite form “puzzle” autors wittily applieded onto the tourney’s theme. I wonder how they would manage with two sixes…

Comment:
A reduced board 5×5 author still more minimized by using of holes (marked by darkgreen color). Squares of that curtailed boards form symbol of fives in every twin.
Known elements joined in new combination. Very graceful.

Comment:
White knightrider will exchange departure squares with black knightrider in position a), or with black bishop in position b). Pieces on the board form symbol 55, which additionally remains on the board even after solving.
Symbolic problem, symbolic commendation.

Author’s solution:
After first solution, i.e. after 4th of Aprile life jogs to the second 55, sleeping pawn “c2” stands up and goes in the way of life – no very different from that first…
2 identical solutions are because even second 55 after 4th of Aprile to look like of that first!
If you dont like sleeping pawns “a1, c1”, you have there 2 nice white promoted queens – together white pieces are also 5!
If you would like change anything in your life, only replace pawns “a2, c2” by substituted pawns from “a1, c1”. Entranced new pawns not change the solution, because destiny of life is given and goes in its predestinated road just on and on…

Comment:
This problem is possible composedly assign to category “conversational chess”, eventually “literary chess”, because to understanding is needed tell one’s story.
Email was sended additionally on 1st of April at the time 05.55, let alone of those “5” at full blast…
Chess unassuming but in relation to the comment it is interesting story.